Question

The solution set of the inequality $max\{1-x^2,|x-1|\}<1$

• A. $(-\infty ,0)\cup (1,\infty )$
• B. $(-\infty ,0)\cup (2,\infty )$
• C. $(0,2)$
• D. $(-1,1)$

Answer 1

The correct option is C $(0,2)$

$max\{1-x^2,|x-1|\}<1$

for $x<-1$ ,

at $x=-2$

$\Rightarrow max\{1-4,|-2-1|\}=max\{-3,3\}=3\nless 1$

for $0>x>-1,$ at $x=\frac{-1}{2}$,

$\Rightarrow max\{1-\frac{1}{4},|-12-1|\}=max\{\frac{3}{4},\frac{1}{2}\}\nless 1$

for , $1>x>0$, at $x=\frac{1}{2}\Rightarrow max\{1-\frac{1}{4},|\frac{1}{2}-1|\}=max\{\frac{3}{4},\frac{1}{2}\}=\frac{3}{4}<1$

for $x>1$, at $x=2,\Rightarrow max\{1-4,|2-1|\}=1\nless 1$

at $x=1,max\{0,\left|0\right|\}=0<1$

for $1<x<2,$at $x=\frac{3}{2}\Rightarrow max\{1-\frac{9}{4},|\frac{3}{2}-1|\}=\frac{1}{2}<1$

$\Rightarrow x\in (0,2)$